# The orbital angular momentum for an electron revolving in an orbit is given by $\dpi{100} \sqrt{l\left ( l+1 \right )}.\frac{h}{2\pi }$This momentum for an s- electron will be given by Option 1) $+\frac{1}{2}.\frac{h}{2\pi }$ Option 2) $zero$ Option 3) $\frac{h}{2\pi }$ Option 4) $\sqrt{2}\cdot \frac{h}{2\pi }$

As we learnt in

Azimuthal Quantum Number(l) -

For a given value of n, l can have n values ranging from 0 to n – 1, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ....( n –1)

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The value of $l$ ( azimuthal quantum number ) for s-electron is equal to zero.

$Orbital\: angular \: momentum\: = \sqrt{l\left ( l+1 \right )\cdot \frac{h}{2\pi }}$

Substituting the value of  $l$ for s-electron $= \sqrt{0\left ( 0+1 \right )}\cdot \frac{h}{2\pi }= 0$

Option 1)

$+\frac{1}{2}.\frac{h}{2\pi }$

Incorrect option

Option 2)

$zero$

Correct Option

Option 3)

$\frac{h}{2\pi }$

Incorrect option

Option 4)

$\sqrt{2}\cdot \frac{h}{2\pi }$

Incorrect option

N

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